Integrals of irrational functions. Integration of irrational functions

Let's remember our happy school years. Pioneers in mathematics lessons, when starting to study roots, first of all became acquainted with the square root. We will go the same way.

Example 1

Find the indefinite integral

Analyzing the integrand, you come to the sad conclusion that it does not at all resemble table integrals. Now, if all this stuff was in the numerator, it would be simple. Or there would be no root below. Or a polynomial. None fraction integration methods They don't help either. What to do?

The main technique for solving irrational integrals is a change of variable, which will rid us of ALL roots in the integrand.

Note that this replacement is a little peculiar; its technical implementation differs from the “classical” replacement method, which was discussed in the lesson Substitution method in indefinite integral.

In this example, you need to replace x = t 2, that is, instead of the “X” under the root we will have t 2. Why the replacement? Because, and as a result of the replacement, the root will disappear.

If instead of the square root we had in the integrand, then we would have made the substitution. If it had been there, they would have carried it out and so on.

Okay, we will turn into . What happens to the polynomial? There are no difficulties: if , then .

It remains to be seen what the differential will turn into. This is done like this:

We take our replacement and hang differentials on both parts:

(we will describe it in as much detail as possible).

The solution format should look something like this:

Let's replace: .

(1) We carry out the substitution after the replacement (how, what and where is already considered).

(2) We take the constant outside the integral. The numerator and denominator are reduced by t.

(3) The resulting integral is tabular; we prepare it for integration by selecting the square.

(4) Integrate over the table using the formula

(5) We carry out the reverse replacement. How it's done? We remember why we danced: if, then.

Example 2

Find the indefinite integral

This is an example for you to solve on your own. Full solution and answer at the end of the lesson.

Somehow it happened that in Examples 1, 2 there is a “bare” numerator with a lonely differential. Let's fix the situation.

Example 3

Find the indefinite integral

A preliminary analysis of the integrand again shows that there is no easy way. And therefore you need to get rid of the root.

Let's replace: .

Behind we denote the ENTIRE expression under the root. The replacement from the previous examples is not suitable here (more precisely, it can be done, but it will not get rid of the root).

We hang differentials on both parts:

We've sorted out the numerator. What to do with in the denominator?

We take our replacement and express from it: .

If, then.

(1) We carry out the substitution in accordance with the replacement performed.

(2) Comb the numerator. Here I chose not to take the constant out of the integral sign (you can do it this way, it won’t be an error)

(3) We expand the numerator into a sum. Once again, we strongly recommend that you read the first paragraph of the lesson Integrating Some Fractions. There will be plenty of rigmarole with decomposing the numerator into a sum in irrational integrals; it is very important to practice this technique.

(4) Divide the numerator by the denominator term by term.

(5) We use the linearity properties of the indefinite integral. In the second integral we select a square for subsequent integration according to the table.

(6) We integrate according to the table. The first integral is quite simple, in the second we use the tabular formula of the high logarithm .

(7) We carry out the reverse replacement. If we carried out a replacement, then back: .

Example 4

Find the indefinite integral

This is an example for you to solve on your own; if you have not carefully worked through the previous examples, you will make a mistake! Full solution and answer at the end of the lesson.

In principle, integrals with several identical roots, for example

Etc. What to do if the integrand has roots different?

Example 5

Find the indefinite integral

Here comes the reckoning for the bare numerators. When such an integral is encountered, it usually becomes scary. But fears are in vain; after making a suitable replacement, the integrand becomes simpler. The task is the following: to carry out a successful replacement in order to immediately get rid of ALL roots.

When given different roots, it is convenient to adhere to a certain solution scheme.

First, we write out the integrand function on a draft, and present all the roots in the form:

We will be interested denominators degrees:

An irrational function of a variable is a function that is formed from a variable and arbitrary constants using a finite number of operations of addition, subtraction, multiplication (raising to an integer power), division and taking roots. An irrational function differs from a rational one in that the irrational function contains operations for extracting roots.

There are three main types of irrational functions, the indefinite integrals of which are reduced to integrals of rational functions. These are integrals containing roots of arbitrary integer powers from a linear fractional function (the roots can be of different powers, but from the same linear fractional function); integrals of a differential binomial and integrals with the square root of a square trinomial.

Important note. Roots have multiple meanings!

When calculating integrals containing roots, expressions of the form are often encountered, where is some function of the integration variable. It should be borne in mind that. That is, at t >< 0 , |t| = t. At t 0 0 , |t| = - t .< 0 Therefore, when calculating such integrals, it is necessary to separately consider the cases t > 0 and t< 0 .

This can be done by writing signs or wherever necessary. Assuming that the top sign refers to the case t >

, and the lower one - to the case t

.
,
With further transformation, these signs, as a rule, cancel each other.
A second approach is also possible, in which the integrand and the result of integration can be considered as complex functions of complex variables. Then you don’t have to pay attention to the signs in radical expressions. This approach is applicable if the integrand is analytic, that is, a differentiable function of a complex variable. In this case, both the integrand and its integral are multivalued functions. Therefore, after integration, when substituting numerical values, it is necessary to select a single-valued branch (Riemann surface) of the integrand, and for it select the corresponding branch of the integration result.
Fractional linear irrationality

These are integrals with roots from the same fractional linear function: where R is a rational function, are rational numbers, m 1, n 1, ..., m s, n s are integers, α, β, γ, δ are real numbers. Such integrals are reduced to the integral of a rational function by substitution: , where n is the common denominator of the numbers r 1, ..., r s.).

The roots may not necessarily come from a linear fractional function, but also from a linear one (γ =
, .

0 , δ = 1

), or on the integration variable x (α =
,
1, β = 0, γ = 0, δ = 1
Here are examples of such integrals:

1) If p is an integer. Substitution x = t N, where N is the common denominator of the fractions m and n.
2) If - an integer. Substitution a x n + b = t M, where M is the denominator of the number p.
3) If - an integer. Substitution a + b x - n = t M, where M is the denominator of the number p.

In other cases, such integrals are not expressed through elementary functions.

Sometimes such integrals can be simplified using reduction formulas:
;
.

Integrals containing the square root of a square trinomial

Such integrals have the form:
,
where R is a rational function. For each such integral there are several methods for solving it.
1) Using transformations lead to simpler integrals.
2) Apply trigonometric or hyperbolic substitutions.
3) Apply Euler substitutions.

Let's look at these methods in more detail.

1) Transformation of the integrand function

Applying the formula and performing algebraic transformations, we reduce the integrand function to the form:
,
where φ(x), ω(x) are rational functions.

Type I

Integral of the form:
,
where P n (x) is a polynomial of degree n.

Such integrals are found by the method of indefinite coefficients using the identity:

.
Differentiating this equation and equating the left and right sides, we find the coefficients A i.

Type II

Integral of the form:
,
where P m (x) is a polynomial of degree m.

Substitution t = (x - α) -1 this integral is reduced to the previous type. If m ≥ n, then the fraction should have an integer part.

III type

Here we do the substitution:
.
After which the integral will take the form:
.
Next, the constants α, β must be chosen such that the coefficients of t in the denominator become zero:
B = 0, B 1 = 0.
Then the integral decomposes into the sum of integrals of two types:
,
,
which are integrated by substitutions:
u 2 = A 1 t 2 + C 1,
v 2 = A 1 + C 1 t -2 .

2) Trigonometric and hyperbolic substitutions

For integrals of the form , a > 0 ,
we have three main substitutions:
;
;
;

For integrals, a > 0 ,
we have the following substitutions:
;
;
;

And finally, for the integrals, a > 0 ,
the substitutions are as follows:
;
;
;

3) Euler substitutions

Also, integrals can be reduced to integrals of rational functions of one of three Euler substitutions:
, for a > 0;
, for c > 0 ;
, where x 1 is the root of the equation a x 2 + b x + c = 0.

If this equation has real roots.

Elliptic integrals
,
where R is a rational function, .

Such integrals are called elliptic. In general, they are not expressed through elementary functions. However, there are cases when there are relationships between the coefficients A, B, C, D, E, in which such integrals are expressed through elementary functions.
.

Below is an example related to reflexive polynomials. The calculation of such integrals is performed using substitutions:

Example
.

Calculate the integral:

Solution

.
Let's make a substitution. 0 Here at x > 0 (u>< 0 ) take the upper sign ′+ ′. At x< 0 (u

) - lower ′- ′.

Answer
References:

This section will discuss the method of integrating rational functions. 7.1. Brief information about rational functions The simplest rational function is a polynomial of the tith degree, i.e. a function of the form where are real constants, and a0 Ф 0. The polynomial Qn(x) whose coefficient a0 = 1 is called reduced. A real number b is called the root of the polynomial Qn(z) if Q„(b) = 0. It is known that each polynomial Qn(x) with real coefficients is uniquely decomposed into real factors of the form where p, q are real coefficients, and the quadratic factors have no real roots and, therefore, cannot be decomposable into real linear factors. By combining identical factors (if any) and assuming, for simplicity, that the polynomial Qn(x) is reduced, we can write its factorization in the form where are natural numbers. Since the degree of the polynomial Qn(x) is equal to n, then the sum of all exponents a, /3,..., A, added to the double sum of all exponents ω,..., q, is equal to n: The root a of a polynomial is called simple or single , if a = 1, and multiple if a > 1; the number a is called the multiplicity of the root a. The same applies to other roots of the polynomial. A rational function f(x) or a rational fraction is the ratio of two polynomials, and it is assumed that the polynomials Pm(x) and Qn(x) do not have common factors. A rational fraction is called proper if the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, i.e. If m n, then the rational fraction is called an improper fraction, and in this case, dividing the numerator by the denominator according to the rule for dividing polynomials, it can be represented in the form where are some polynomials, and ^^ is a proper rational fraction. Example 1. A rational fraction is an improper fraction. Dividing by a “corner”, we have Therefore. Here. and it's a proper fraction. To find these constants, the right-hand side of equality (I) is brought to a common denominator, and then the coefficients at the same powers of x in the numerators of the left and right sides are equated. This gives a system of linear equations from which the required constants are found. . This method of finding unknown constants is called the method of undetermined coefficients. Sometimes it is more convenient to use another method of finding unknown constants, which consists in the fact that after equating the numerators, an identity is obtained with respect to x, in which the argument x is given some values, for example, the values of the roots, resulting in equations for finding the constants. It is especially convenient if the denominator Q„(x) has only real simple roots. Example 2. Decompose the rational fraction into simpler fractions. This fraction is proper. We decompose the denominator into multiplies: Since the roots of the denominator are real and different, then, based on formula (1), the decomposition of the fraction into the simplest will have the form: Reducing the right honor “of that equality to the common denominator and equating the numerators on its left and right sides, we obtain the identity or We find unknown coefficients A. 2?, C in two ways. First way Equating the coefficients for the same powers of x, t.v. with (free term), and the left and right sides of the identity, we obtain a linear system of equations for finding the unknown coefficients A, B, C: This system has a unique solution C The second method. Since the roots of the denominator are torn at i 0, we get 2 = 2A, whence A * 1; g i 1, we get -1 * -B, from which 5 * 1; x i 2, we get 2 = 2C. whence C» 1, and the required expansion has the form 3. Rehlozhnt not the simplest fractions rational fraction 4 We decompose the polynomial, which is in the opposite direction, into factors: . The denominator has two different real roots: x\ = 0 multiplicity of multiplicity 3. Therefore, the decomposition of this fraction is not the simplest: Reducing the right-hand side to a common denominator, we find or The first method. Equating the coefficients for the same powers of x in the left and right sides of the last identity. we obtain a linear system of equations. This system has a unique solution and the required expansion will be the Second method. In the resulting identity, putting x = 0, we obtain 1 a A2, or A2 = 1; field* gay x = -1, we get -3 i B), or Bj i -3. When substituting the found values of the coefficients A\ and B) and the identity will take the form or Putting x = 0, and then x = -I. we find that = 0, B2 = 0 and. this means B\ = 0. Thus, we again obtain Example 4. Expand the rational fraction 4 into simpler fractions. The denominator of the fraction has no real roots, since the function x2 + 1 does not vanish for any real values of x. Therefore, the decomposition into simple fractions should have the form From here we get or. Equating the coefficients of the synax powers of x in the left and right sides of the last equality, we will have where we find and, therefore, It should be noted that in some cases decompositions into simple fractions can be obtained faster and easier by acting in some other way, without using the method of indefinite coefficients For example, to obtain the decomposition of the fraction in example 3, you can add and subtract in the numerator 3x2 and divide as indicated below. 7.2. Integration of simple fractions, As mentioned above, any improper rational fraction can be represented as the sum of some polynomial and a proper rational fraction (§7), and this representation is unique. Integrating a polynomial is not difficult, so consider the question of integrating a proper rational fraction. Since any proper rational fraction can be represented as a sum of simple fractions, its integration is reduced to the integration of simple fractions. Let us now consider the question of their integration. III. To find the integral of the simplest fraction of the third type, we isolate the complete square of the binomial from the square trinomial: Since the second term is equal to a2, where and then we make the substitution. Then, taking into account the linear properties of the integral, we find: Example 5. Find the integral 4 The integrand function is the simplest fraction of the third type, since the square trinomial x1 + Ax + 6 has no real roots (its discriminant is negative: , and the numerator contains a polynomial of the first degree. Therefore, we proceed as follows: 1) select a perfect square in the denominator 2) make a substitution (here 3) by * one integral To find the integral of the simplest fraction of the fourth type, we put, as above, . Then we get the Integral on the right side denoted by A and transform it as follows: Integral on the right side is integrated by parts, assuming from where or Integration of rational functions Brief information about rational functions Integration of simple fractions General case Integration of irrational functions Euler's first substitution Second Euler substitution Third substitution Euler We have obtained the so-called recurrent formula, which allows us to find the integral Jk for any k = 2, 3,. .. . Indeed, the integral J\ is tabular: Putting in the recurrence formula, we find Knowing and putting A = 3, we can easily find Jj and so on. In the final result, substituting everywhere instead of t and a their expressions in terms of x and coefficients p and q, we obtain for the initial integral its expression in terms of x and the given numbers M, LG, p, q. Example 8. New integral “The integrand function is the simplest fraction of the fourth type, since the discriminant of a square trinomial is negative, i.e. This means that the denominator has no real roots, and the numerator is a polynomial of the 1st degree. 1) We select a complete square in the denominator 2) We make a substitution: The integral will take the form: Putting in the recurrence formula * = 2, a3 = 1. we will have, and, therefore, the desired integral is equal Returning to the variable x, we finally obtain 7.3. General case From the results of paragraphs. 1 and 2 of this section immediately follows an important theorem. Theorem! 4. The indefinite integral of any rational function always exists (on intervals in which the denominator of the fraction Q„(x) φ 0) and is expressed through a finite number of elementary functions, namely, it is an algebraic sum, the terms of which can only be multiplied , rational fractions, natural logarithms and arctangents. So, to find the indefinite integral of a fractional-rational function, one should proceed in the following way: 1) if the rational fraction is improper, then by dividing the numerator by the denominator, the whole part is isolated, i.e., this function is represented as the sum of a polynomial and a proper rational fraction; 2) then the denominator of the resulting proper fraction is decomposed into the product of linear and quadratic factors; 3) this proper fraction is decomposed into the sum of simple fractions; 4) using the linearity of the integral and the formulas of step 2, the integrals of each term are found separately. Example 7. Find the integral M Since the denominator is a polynomial of the third order, the integrand function is an improper fraction. We highlight the whole part in it: Therefore, we will have. The denominator of a proper fraction has phi distinct real roots: and therefore its decomposition into simple fractions has the form Hence we find. Giving the argument x values equal to the roots of the denominator, we find from this identity that: Therefore, the required integral will be equal to Example 8. Find the integral 4 The integrand is a proper fraction, the denominator of which has two different real roots: x - O multiplicity of 1 and x = 1 of multiplicity 3, Therefore, the expansion of the integrand into simple fractions has the form Bringing the right side of this equality to a common denominator and reducing both sides of the equality by this denominator, we obtain or. We equate the coefficients for the same powers of x on the left and right sides of this identity: From here we find. Substituting the found values of the coefficients into the expansion, we will have. Integrating, we find: Example 9. Find the integral 4 The denominator of the fraction has no real roots. Therefore, the expansion of the integrand into simple fractions has the form Hence or Equating the coefficients for the same powers of x on the left and right sides of this identity, we will have from where we find and, therefore, Remark. In the given example, the integrand function can be represented as a sum of simple fractions in a simpler way, namely, in the numerator of the fraction we select the binary that is in the denominator, and then we perform term-by-term division: §8. Integration of irrational functions A function of the form where Pm and £?„ are polynomials of degree type, respectively, in the variables uub2,... is called a rational function of ubu2j... For example, a polynomial of the second degree in two variables u\ and u2 has the form where - some real constants, and Example 1, The function is a rational function of the variables r and y, since it represents the ratio of a polynomial of the third degree and a polynomial of the fifth degree, but is not a yew function. In the case when the variables, in turn, are functions of the variable x: then the function ] is called a rational function of the functions of the Example. A function is a rational function of r and rvdikvlv Pryaivr 3. A function of the form is not a rational function of x and the radical y/r1 + 1, but it is a rational function of functions. As examples show, integrals of irrational functions are not always expressed through elementary functions. For example, integrals often encountered in applications are not expressed in terms of elementary functions; these integrals are called elliptic integrals of the first and second kind, respectively. Let us consider those cases when the integration of irrational functions can be reduced, with the help of some substitutions, to the integration of rational functions. 1. Let it be necessary to find the integral where R(x, y) is a rational function of its arguments x and y; m £ 2 - natural number; a, 6, c, d are real constants that satisfy the condition ad - bc ^ O (for ad - be = 0, the coefficients a and b are proportional to the coefficients c and d, and therefore the relationship does not depend on x; this means that in this case the integrand function will be a rational function of the variable x, the integration of which was discussed earlier). Next we find or, after simplification, Therefore where A1 (t) is a rational function of *, since the rational funadia of a rational function, as well as the product of rational functions, are rational functions. We know how to integrate rational functions. Let Then the required integral be equal to At. IvYti integral 4 An integrand* function is a rational function of. Therefore, we set t = Then Integration of rational functions Brief information about rational functions Integration of simple fractions General case Integration of irrational functions Euler's first substitution Euler's second substitution Euler's third substitution Thus, we obtain Primar 5. Find the integral The common denominator of fractional exponents of x is equal to 12, so the integrand the function can be represented in the form 1 _ 1_ which shows that it is a rational function of: Taking this into account, let us put. Consequently, 2. Consider intephs of the form where the subintephal function is such that by replacing the radical \/ax2 + bx + c in it by y, we obtain a function R(x) y) - rational with respect to both arguments x and y. This integral is reduced to the integral of a rational function of another variable using Euler's substitutions. 8.1. Euler's first substitution Let the coefficient a > 0. Let us set or Hence we find x as a rational function of u, which means Thus, the indicated substitution expresses rationally in terms of *. Therefore, we will have a remark. The first Euler substitution can also be taken in the form Example 6. Let’s find the integral Therefore, we will have dx Euler’s substitution, show that Y 8.2. Euler's second substitution Let the trinomial ax2 + bx + c have different real roots R] and x2 (the coefficient can have any sign). In this case, we assume Since then we obtain Since x,dxn y/ax2 + be + c are expressed rationally in terms of t, then the original integral is reduced to the integral of a rational function, i.e. where Problem. Using Euler's first substitution, show that is a rational function of t. Example 7. Find the integral dx M function ] - x1 has different real roots. Therefore, we apply the second Euler substitution. From here we find. Substituting the found expressions into the Given?v*gyvl; we get 8.3. Third Euler substascom Let the coefficient c > 0. We make a change of variable by putting. Note that to reduce the integral to the integral of a rational function, the first and second Euler substitutions are sufficient. In fact, if the discriminant b2 -4ac > 0, then the roots of the quadratic trinomial ax + bx + c are real, and in this case the second Euler substitution is applicable. If, then the sign of the trinomial ax2 + bx + c coincides with the sign of the coefficient a, and since the trinomial must be positive, then a > 0. In this case, Euler’s first substitution is applicable. To find integrals of the type indicated above, it is not always advisable to use Euler’s substitutions, since for them it is possible to find other methods of integration that lead to the goal faster. Let's consider some of these integrals. 1. To find integrals of the form, isolate the perfect square from the square of the th trinomial: where After this, make a substitution and get where the coefficients a and P have different signs or they are both positive. For, and also for a > 0, the integral will be reduced to a logarithm, and if so, to the arcsine. At. Find imtegral 4 Sokak then. Assuming, we get Prmmar 9. Find. Assuming x -, we will have 2. The integral of the form is reduced to the integral y from step 1 as follows. Considering that the derivative ()" = 2, we highlight it in the numerator: 4 We identify the derivative of the radical expression in the numerator. Since (x, then we will have, taking into account the result of example 9, 3. Integrals of the form where P„(x) is a polynomial n -th degree, can be found by the method of indefinite coefficients, which consists of the following. Let us assume that the equality is Example 10. Mighty integral where Qn-i(s) is a polynomial of (n - 1) degree with indefinite coefficients: To find unknowns. coefficients | we differentiate both sides of (1): Then we reduce the right side of equality (2) to a common denominator equal to the denominator of the left side, i.e. y/ax2 + bx + c, reducing both sides of (2) by which we obtain the identity in both sides of which contain polynomials of degree n. Equating the coefficients for the same degrees of x in the left and right sides of (3), we obtain n + 1 equations, from which we find the required coefficients j4*(fc = 0,1,2,..., n ). Substituting their values into the right side of (1) and finding the integral + c we obtain the answer for this integral. Example 11. Find the integral Let's put Differentiating both suits of the equality, we will have Bringing the right side to a common denominator and reducing both sides by it, we get the identity or. Equating the coefficients at the same powers of x, we arrive at a system of equations from which we find = Then we find the integral on the right side of equality (4): Consequently, the required integral will be equal to

N.M. Gunter, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003. Under irrational understand an expression in which the independent variable %%x%% or the polynomial %%P_n(x)%% of degree %%n \in \mathbb(N)%% is included under the sign radical (from Latin radix

- root), i.e. raised to a fractional power. By replacing a variable, some classes of integrands that are irrational with respect to %%x%% can be reduced to rational expressions with respect to a new variable.

The concept of a rational function of one variable can be extended to multiple arguments. If for each argument %%u, v, \dotsc, w%% when calculating the value of a function, only arithmetic operations and raising to an integer power are provided, then we speak of a rational function of these arguments, which is usually denoted %%R(u, v, \ dotsc, w)%%. The arguments of such a function can themselves be functions of the independent variable %%x%%, including radicals of the form %%\sqrt[n](x), n \in \mathbb(N)%%. For example, the rational function $$ R(u,v,w) = \frac(u + v^2)(w) $$ with %%u = x, v = \sqrt(x)%% and %%w = \sqrt(x^2 + 1)%% is a rational function of $$ R\left(x, \sqrt(x), \sqrt(x^2+1)\right) = \frac(x + \sqrt(x ^2))(\sqrt(x^2 + 1)) = f(x) $$ from %%x%% and radicals %%\sqrt(x)%% and %%\sqrt(x^2 + 1 )%%, while the function %%f(x)%% will be an irrational (algebraic) function of one independent variable %%x%%.

Let's consider integrals of the form %%\int R(x, \sqrt[n](x)) \mathrm(d)x%%. Such integrals are rationalized by replacing the variable %%t = \sqrt[n](x)%%, then %%x = t^n, \mathrm(d)x = nt^(n-1)%%.

Example 1

The integrand of the desired argument is written as a function of radicals of degree %%2%% and %%3%%. Since the least common multiple of %%2%% and %%3%% is %%6%%, this integral is an integral of type %%\int R(x, \sqrt(x)) \mathrm(d)x %% and can be rationalized by replacing %%\sqrt(x) = t%%. Then %%x = t^6, \mathrm(d)x = 6t \mathrm(d)t, \sqrt(x) = t^3, \sqrt(x) =t^2%%. Therefore, $$ \int \frac(\mathrm(d)x)(\sqrt(x) + \sqrt(x)) = \int \frac(6t^5 \mathrm(d)t)(t^3 + t^2) = 6\int\frac(t^3)(t+1)\mathrm(d)t. $$ Let's take %%t + 1 = z, \mathrm(d)t = \mathrm(d)z, z = t + 1 = \sqrt(x) + 1%% and $$ \begin(array)(ll ) \int \frac(\mathrm(d)x)(\sqrt(x) + \sqrt(x)) &= 6\int\frac((z-1)^3)(z) \mathrm(d) t = \\ &= 6\int z^2 dz -18 \int z \mathrm(d)z + 18\int \mathrm(d)z -6\int\frac(\mathrm(d)z)(z ) = \\ &= 2z^3 - 9 z^2 + 18z -6\ln|z| + C = \\ &= 2 \left(\sqrt(x) + 1\right)^3 - 9 \left(\sqrt(x) + 1\right)^2 + \\ &+~ 18 \left( \sqrt(x) + 1\right) - 6 \ln\left|\sqrt(x) + 1\right| + C \end(array) $$

Integrals of the form %%\int R(x, \sqrt[n](x)) \mathrm(d)x%% are a special case of fractional linear irrationalities, i.e. integrals of the form %%\displaystyle\int R\left(x, \sqrt[n](\dfrac(ax+b)(cd+d))\right) \mathrm(d)x%%, where %%ad - bc \neq 0%%, which can be rationalized by replacing the variable %%t = \sqrt[n](\dfrac(ax+b)(cd+d))%%, then %%x = \dfrac(dt^n - b)(a - ct^n)%%. Then $$ \mathrm(d)x = \frac(n t^(n-1)(ad - bc))(\left(a - ct^n\right)^2)\mathrm(d)t. $$

Example 2

Find %%\displaystyle\int \sqrt(\dfrac(1 -x)(1 + x))\dfrac(\mathrm(d)x)(x + 1)%%.

Let's take %%t = \sqrt(\dfrac(1 -x)(1 + x))%%, then %%x = \dfrac(1 - t^2)(1 + t^2)%%, $$ \begin(array)(l) \mathrm(d)x = -\frac(4t\mathrm(d)t)(\left(1 + t^2\right)^2), \\ 1 + x = \ frac(2)(1 + t^2), \\ \frac(1)(x + 1) = \frac(1 + t^2)(2). \end(array) $$ Therefore, $$ \begin(array)(l) \int \sqrt(\dfrac(1 -x)(1 + x))\frac(\mathrm(d)x)(x + 1) = \\ = \frac(t(1 + t^2))(2)\left(-\frac(4t \mathrm(d)t)(\left(1 + t^2\right)^2 )\right) = \\ = -2\int \frac(t^2\mathrm(d)t)(1 + t^2) = \\ = -2\int \mathrm(d)t + 2\int \frac(\mathrm(d)t)(1 + t^2) = \\ = -2t + \text(arctg)~t + C = \\ = -2\sqrt(\dfrac(1 -x)( 1 + x)) + \text(arctg)~\sqrt(\dfrac(1 -x)(1 + x)) + C. \end(array) $$

Let's consider integrals of the form %%\int R\left(x, \sqrt(ax^2 + bx + c)\right) \mathrm(d)x%%. In the simplest cases, such integrals are reduced to tabular ones if, after isolating the complete square, a change of variables is made.

Example 3

Find the integral %%\displaystyle\int \dfrac(\mathrm(d)x)(\sqrt(x^2 + 4x + 5))%%.

Considering that %%x^2 + 4x + 5 = (x+2)^2 + 1%%, we take %%t = x + 2, \mathrm(d)x = \mathrm(d)t%%, then $$ \begin(array)(ll) \int \frac(\mathrm(d)x)(\sqrt(x^2 + 4x + 5)) &= \int \frac(\mathrm(d)t) (\sqrt(t^2 + 1)) = \\ &= \ln\left|t + \sqrt(t^2 + 1)\right| + C = \\ &= \ln\left|x + 2 + \sqrt(x^2 + 4x + 5)\right| + C. \end(array) $$

In more complex cases, to find integrals of the form %%\int R\left(x, \sqrt(ax^2 + bx + c)\right) \mathrm(d)x%% are used

There is no universal way to solve irrational equations, since their class differs in quantity. The article will highlight characteristic types of equations with substitution using the integration method.

To use the direct integration method, it is necessary to calculate indefinite integrals of the type ∫ k x + b p d x , where p is a rational fraction, k and b are real coefficients.

Example 1

Find and calculate the antiderivatives of the function y = 1 3 x - 1 3 .

Solution

According to the integration rule, it is necessary to apply the formula ∫ f (k x + b) d x = 1 k F (k x + b) + C, and the table of antiderivatives indicates that there is a ready-made solution to this function. We get that

∫ d x 3 x - 1 3 = ∫ (3 x - 1) - 1 3 d x = 1 3 1 - 1 3 + 1 (3 x - 1) - 1 3 + 1 + C = = 1 2 (3 x - 1) 2 3 + C

Answer:∫ d x 3 x - 1 3 = 1 2 (3 x - 1) 2 3 + C .

There are cases when it is possible to use the method of subsuming a differential sign. This is solved by the principle of finding indefinite integrals of the form ∫ f " (x) · (f (x)) p d x , when the value of p is considered a rational fraction.

Example 2

Find the indefinite integral ∫ 3 x 2 + 5 x 3 + 5 x - 7 7 6 d x .

Solution

Note that d x 3 + 5 x - 7 = x 3 + 5 x - 7 "d x = (3 x 2 + 5) d x. Then it is necessary to subsume the differential sign using tables of antiderivatives. We obtain that

∫ 3 x 2 + 5 x 3 + 5 x - 7 7 6 d x = ∫ (x 3 + 5 x - 7) - 7 6 (3 x 2 + 5) d x = = ∫ (x 3 + 5 x - 7 ) - 7 6 d (x 3 + 5 x - 7) = x 3 + 5 x - 7 = z = = ∫ z - 7 6 d z = 1 - 7 6 + 1 z - 7 6 + 1 + C = - 6 z - 1 6 + C = z = x 3 + 5 x - 7 = - 6 (x 3 + 5 x - 7) 6 + C

Answer:∫ 3 x 2 + 5 x 3 + 5 x - 7 7 6 d x = - 6 (x 3 + 5 x - 7) 6 + C .

Solving indefinite integrals involves a formula of the form ∫ d x x 2 + p x + q, where p and q are real coefficients. Then you need to select a complete square from under the root. We get that

x 2 + p x + q = x 2 + p x + p 2 2 - p 2 2 + q = x + p 2 2 + 4 q - p 2 4

Applying the formula located in the table of indefinite integrals, we obtain:

∫ d x x 2 ± α = ln x + x 2 ± α + C

Then the integral is calculated:

∫ d x x 2 + p x + q = ∫ d x x + p 2 2 + 4 q - p 2 4 = = ln x + p 2 + x + p 2 2 + 4 q - p 2 4 + C = = ln x + p 2 + x 2 + p x + q + C

Example 3

Find the indefinite integral of the form ∫ d x 2 x 2 + 3 x - 1 .

Solution

To calculate, you need to take out the number 2 and place it in front of the radical:

∫ d x 2 x 2 + 3 x - 1 = ∫ d x 2 x 2 + 3 2 x - 1 2 = 1 2 ∫ d x x 2 + 3 2 x - 1 2

Select a complete square in radical expression. We get that

x 2 + 3 2 x - 1 2 = x 2 + 3 2 x + 3 4 2 - 3 4 2 - 1 2 = x + 3 4 2 - 17 16

Then we obtain an indefinite integral of the form 1 2 ∫ d x x 2 + 3 2 x - 1 2 = 1 2 ∫ d x x + 3 4 2 - 17 16 = = 1 2 ln x + 3 4 + x 2 + 3 2 x - 1 2 + C

Answer: d x x 2 + 3 x - 1 = 1 2 ln x + 3 4 + x 2 + 3 2 x - 1 2 + C

Integration of irrational functions is carried out in a similar way. Applicable for functions of the form y = 1 - x 2 + p x + q.

Example 4

Find the indefinite integral ∫ d x - x 2 + 4 x + 5 .

Solution

First you need to derive the square of the denominator of the expression from under the root.

∫ d x - x 2 + 4 x + 5 = ∫ d x - x 2 - 4 x - 5 = = ∫ d x - x 2 - 4 x + 4 - 4 - 5 = ∫ d x - x - 2 2 - 9 = ∫ d x - (x - 2) 2 + 9

The table integral has the form ∫ d x a 2 - x 2 = a r c sin x a + C, then we obtain that ∫ d x - x 2 + 4 x + 5 = ∫ d x - (x - 2) 2 + 9 = a r c sin x - 2 3 +C

Answer:∫ d x - x 2 + 4 x + 5 = a r c sin x - 2 3 + C .

The process of finding antiderivative irrational functions of the form y = M x + N x 2 + p x + q, where the existing M, N, p, q are real coefficients, and are similar to the integration of simple fractions of the third type. This transformation has several stages:

summing the differential under the root, isolating the complete square of the expression under the root, using tabular formulas.

Example 5

Find the antiderivatives of the function y = x + 2 x 2 - 3 x + 1.

Solution

From the condition we have that d (x 2 - 3 x + 1) = (2 x - 3) d x and x + 2 = 1 2 (2 x - 3) + 7 2, then (x + 2) d x = 1 2 (2 x - 3) + 7 2 d x = 1 2 d (x 2 - 3 x + 1) + 7 2 d x .

Let's calculate the integral: ∫ x + 2 x 2 - 3 x + 1 d x = 1 2 ∫ d (x 2 - 3 x + 1) x 2 - 3 x + 1 + 7 2 ∫ d x x 2 - 3 x + 1 = = 1 2 ∫ (x 2 - 3 x + 1) - 1 2 d (x 2 - 3 x + 1) + 7 2 ∫ d x x - 3 2 2 - 5 4 = = 1 2 1 - 1 2 + 1 x 2 - 3 x + 1 - 1 2 + 1 + 7 2 ln x - 3 2 + x - 3 2 - 5 4 + C = = x 2 - 3 x + 1 + 7 2 ln x - 3 2 + x 2 - 3 x + 1 + C

Answer:∫ x + 2 x 2 - 3 x + 1 d x = x 2 - 3 x + 1 + 7 2 ln x - 3 2 + x 2 - 3 x + 1 + C .

The search for indefinite integrals of the function ∫ x m (a + b x n) p d x is carried out using the substitution method.

To solve it is necessary to introduce new variables:

When p is an integer, then x = z N is considered, and N is the common denominator for m, n.
When m + 1 n is an integer, then a + b x n = z N, and N is the denominator of p.
When m + 1 n + p is an integer, then the variable a x - n + b = z N is required, and N is the denominator of the number p.

Example 6

Find the definite integral ∫ 1 x 2 x - 9 d x .

Solution

We get that ∫ 1 x 2 x - 9 d x = ∫ x - 1 · (- 9 + 2 x 1) - 1 2 d x . It follows that m = - 1, n = 1, p = - 1 2, then m + 1 n = - 1 + 1 1 = 0 is an integer. You can introduce a new variable of the form - 9 + 2 x = z 2. It is necessary to express x in terms of z. As output we get that

9 + 2 x = z 2 ⇒ x = z 2 + 9 2 ⇒ d x = z 2 + 9 2 " d z = z d z - 9 + 2 x = z

It is necessary to make a substitution into the given integral. We have that

∫ d x x 2 x - 9 = ∫ z d z z 2 + 9 2 z = 2 ∫ d z z 2 + 9 = = 2 3 a r c t g z 3 + C = 2 3 a r c c t g 2 x - 9 3 + C

Answer:∫ d x x 2 x - 9 = 2 3 a r c c t g 2 x - 9 3 + C .

To simplify the solution of irrational equations, basic integration methods are used.

If you notice an error in the text, please highlight it and press Ctrl+Enter